How can I get video duration in seconds?

What I've tried:

ffmpeg -i file.flv 2>&1 | grep "Duration"
  Duration: 00:39:43.08, start: 0.040000, bitrate: 386 kb/s

mediainfo file.flv | grep Duration
Duration : 39mn 43s

This what close, but it's not so accurate, 2383 is 39.71 minutes

ffmpeg -i file.flv 2>&1 | grep "Duration"| cut -d ' ' -f 4 | sed s/,// | sed 's@\..*@@g' | awk '{ split($1, A, ":"); split(A[3], B, "."); print 3600*A[1] + 60*A[2] + B[1] }'

12 Answers 12


To get minutes, you have to divide 2383 seconds by 60.


and then multiply the fractional part .7167 by 60 to get the remaining seconds.


So it's 39 minutes, 43 seconds. The application appears to be giving you an accurate value.

  • 5
    Best answer because you identified the problem in the given solution. Most people would suggest a new solution. Apr 25, 2020 at 14:51

Just use ffprobe directly. No need for sed, grep, etc. There are several "durations" you can acquire (depending on your input).

Format (container) duration

ffprobe -v error -show_entries format=duration -of default=noprint_wrappers=1:nokey=1 input.mp4



Adding the -sexagesimal option will use the HOURS:MM:SS.MICROSECONDS time unit format:


Video stream duration

If you want the duration of a particular video or audio stream:

ffprobe -v error -select_streams v:0 -show_entries stream=duration -of default=noprint_wrappers=1:nokey=1 input.mp4



The above commands are from FFmpeg Wiki: FFprobe Tips.

With ffmpeg

You can use ffmpeg to get duration by decoding the input:

ffmpeg -i input.mp4 -f null -
frame= 1587 fps=0.0 q=0.0 Lsize=N/A time=00:01:03.48 bitrate=N/A

In this example time=00:01:03.48 is the duration.

This may take a long time depending on your input file.

  • These do not work for .m2v files do you have a solution for them aswell
    – utdev
    Jan 5, 2017 at 9:47
  • @utdev See the "With ffmpeg" section.
    – llogan
    Jan 5, 2017 at 18:39
  • Your third solution gives me a wrong time, my video has a length of 01:19:00 but the command returns me time=01:09:15.32, do you have a guess why this is happening
    – utdev
    Feb 15, 2017 at 10:59
  • 1
    For anyone using avprobe, the parameters are slightly different: avprobe -v error -show_format_entry duration .\Sample.mp4
    – Brad
    Jun 18, 2019 at 22:25
  • Certain container types (e.g. Matroska [.mkv]) can be queried for duration at the container level ("format"), but you won't get a result at the video stream level. Sep 22, 2020 at 13:18

If you have ffmpeg, you should also have ffprobe:

ffprobe -i input.file -show_format | grep duration
ffprobe -i input.file -show_format -v quiet | sed -n 's/duration=//p'

This will also give fractions of seconds, if that's a problem you can further process that away with sed.

  • how can I remove the fraction part of seconds?
    – Tina J
    Aug 18, 2017 at 13:42
  • @TinaJ Just pipe it with printf, like | xargs printf %.0f - this will return an integer value.
    – Ilia Ross
    Aug 14, 2018 at 9:21
  • @TinaJ Also if you don't want to round value mathematically but want just to remove decimal part you can assign the duration to a variable duration=$(ffprobe -i input.file -show_format -v quiet | sed -n 's/duration=//p') And remove decimal part echo ${duration%.*}
    – dmitry1100
    May 18, 2020 at 11:18
mediainfo --Output='General;%Duration%' file.flv

This outputs the duration in milliseconds as a single integer value. No need for grep/cut/sed/...

  • Thanks - this was just what I was looking for. Can be installed on MacOS with brew install mediainfo
    – Alex K
    Nov 22, 2018 at 17:32

Solution with mplayer that gives seconds directly:

mplayer -identify -frames 0 -vo null -nosound file.flv 2>&1 | awk -F= '/LENGTH/{print $2}'

For my ffmpeg-0.6.5-1.el6.rf.x86_64, to get just the second, ffprobe command format is:

ffprobe <file> -show_format 2>&1 | sed -n 's/duration=//p' 
  • 1
    how can I remove the fraction part of seconds?
    – Tina J
    Aug 18, 2017 at 13:42
  • To remove fraction part, use: sed -n -E 's/duration=([0-9]+).*/\1/p which extracts the first matching numbers (discarding anything after a non-numeric symbol) using a backreference (\1, which only works with the -E option).
    – Yeti
    Jul 16, 2020 at 12:46

I came across the issue of getting some strange and incorrect metadata from some video files I was working with and I couldn't succeed on finding a pattern or any type of handling using code and tools like ffmpeg, mp4box, ffprobe, mediainfo, mplayer, to get the real duration of the video.

Identifying the real duration of the video was a requirement for a project I was working and the only way I found to get it always right was to reencode the video file using ffmpeg and forcing encoded files to ignore original file's metadata, like:

ffmpeg -i INPUT_FILENAME -acodec copy -vcodec copy -map_metadata -1 OUTPUT_FILENAME"

(This might run faster than you expect. I got surprised, for the type of content and on the environment I was using, average time was 2 seconds)

... and then get duration using a tool at your choice. I like mediainfo, btw: - "mediainfo FILE --Inform="Video;%Duration%" gives you duration in miliseconds.

  • You did not provide more information or better solution than other answers
    – Sam
    Jun 1, 2017 at 7:36
  • No other answers mention -map_metadata -1. This solution was helpful for me
    – jaksco
    Dec 25, 2023 at 6:18


You can extract streams using following code

import json

infile = "<path to your file>"
command = "ffprobe -v quiet -print_format json -show_format  -show_streams {}".format(infile)
res = json.loads(subprocess.check_output(command, shell=True))
streams = res.get('streams')

Each stream would have a field duration.

Special case of Webm

If your input file is webm, Your method of extracting stream duration is going to be different, since the standard store it in a different location. In each stream you would see a field 'tags'

"tags": {
                "language": "eng",
                "HANDLER_NAME": "ISO Media file produced by Google Inc. Created on: 12/09/2019.",
                "VENDOR_ID": "[0][0][0][0]",
                "ENCODER": "Lavc59.4.101 libvorbis",
                "DURATION": "00:03:14.703000000"

You would have to convert this human readable time format into seconds. Which you can easily do with following utility function.

def convert_secs(text):
    if isinstance(text, float):
        num = str(text)
        nums = num.split('.')
        nums = text.split(':')
    if len(nums) == 2:
        st_sn = int(nums[0]) * 60 + float(nums[1])
        return st_sn
    elif len(nums) == 3:
        st_sn = int(nums[0]) * 3600 + int(nums[1]) * 60 + float(nums[2])
        return st_sn
        raise ValueError("Not correct time")
# Returns duration (in seconds) of a video $1 (uses ffmpeg).
get_video_duration() {
  OUTPUT=$(ffmpeg -i "$1" -vframes 1 -f rawvideo -y /dev/null 2>&1) ||
    { debug -e "get_video_duration: error running ffmpeg:\n$OUTPUT"; return 1; }
  DURATION=$(echo "$OUTPUT" | grep -m1 "^[[:space:]]*Duration:" |
    cut -d":" -f2- | cut -d"," -f1 | sed "s/[:\.]/ /g") || 
    { debug -e "get_video_duration: error parsing duration:\n$OUTPUT"; return 1; }
  echo $((10#$HOURS * 3600 + 10#$MINUTES * 60 + 10#$SECONDS))      

Usage:DURATION=$(get_video_duration "$VIDEO")

  • +1 for -vframes 1 but I find -f null -c copy - more portable than -f rawvideo -y /dev/null because /dev/null is problematic on Windows
    – robert4
    Feb 2 at 5:52

I was a bit surprised to find that the simplest way of doing this isn't here, and that the closest thing doesn't give integers, but adds incorrect decimal points that need to be rounded off.

It's quite easy: just use your languages modulus operator.

In php:

$vidLength = 2383; //seconds
$secondsPerMinute = 60;

$seconds = $vidLength % $secondsPerMinute;
//$seconds = 43

$minutes = ($vidLength - $seconds) / $secondsPerMinute;
//$minutes = 39

No rounding errors or weird fractional parts. Just simple, plain math using one of the 6 basic arithmetic operators just like + or - that can be found in any programming language.

Also, the actual perceived error is that you've done whole minutes and then have divided the seconds by 60. 43/60 = 0,7167, effectively turning it from actual seconds to hundredths of a minute.

  • Noone asked for a new solution, not even in PHP.
    – jmizv
    Dec 21, 2023 at 10:18
  • You prefer an overly complicated approach that produces rounding errors? OK, good luck with that. Jan 6 at 9:14

Duration information can appear on several place, for example, mp4 files may have duration under stream sections, but mkv files may not. So the purpose is to print duration information from as many places as possible:

ffprobe.exe  -show_entries format=duration:stream_tags=duration:stream=duration -v quiet -print_format json input_file.mp4

-show_entries is a flag with very weird syntax, the meaning of the above is that to print:

  • duration under format section
  • duration under tags under stream sections
  • duration under stream sections.

-print_format json is optional. JSON is more readable then the default format.

You can also use -sections to see what are the sections in the file and add more to the above to make it more complete.


If you only need to query metadata:

ffprobe -hide_banner -v quiet -show_streams -print_format flat video.mp4




So you can parse it:

while read -r; do
  if [[ "$REPLY" =~ ^streams\.stream\.([0-9])+\.duration=\"([^"]+)\"$ ]]; then
    echo -E Duration of stream "${BASH_REMATCH[1]}": "${BASH_REMATCH[2]}"
done < <(ffprobe -hide_banner -v quiet -show_streams -print_format flat video.mp4)

But if you want to get the effective container's duration, you need to decode it:

AV_LOG_FORCE_NOCOLOR=y ffmpeg -nostdin -hide_banner -nostats -loglevel info -i video.mp4 -f null -c copy - 2>&1 | tail -n 2

It will take some CPU time to decode it, until:


frame=130527 fps=53271 q=-1.0 Lsize=N/A time=01:27:01.12 bitrate=N/A speed=2.13e+03x
video:152014kB audio:40790kB subtitle:0kB other streams:0kB global headers:0kB muxing overhead: unknown

Then, you can parse it:

if [[ "$(AV_LOG_FORCE_NOCOLOR=y ffmpeg -nostdin -hide_banner -nostats -loglevel info -i video.mp4 -f null -c copy - 2>&1 | tail -n 2 | head -n 1)" =~ \ time=([0-9]+):([0-9]{2}):([0-9]{2})\.([0-9]+) ]]; then
  declare duration=0 us="${BASH_REMATCH[4]}" t
  for t in "${BASH_REMATCH[@]:1:3}"; do
    ((duration *= 60))
    ((duration += ${t#0} ))
  while [ ${#us} -lt 6 ]; do us+=0; done
  ((us >= 500000)) && ((duration++))
  ((duration)) || ((duration++))
echo -E Duration: "$duration"

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